Prove that $\lim_{n\rightarrow\infty} \langle S_{k_n}u,x\rangle=\langle u,y\rangle$ using Banach-Alaoglu Throrem

Question

Let $\mathscr{H}$ be separable Hilbert space and let $\{S_n\}\subseteq B(\mathscr{H})$ satisfy $\sup_n\|S_n\|_{\mathscr{H}\rightarrow\mathscr{H}}=M<\infty$. Fix $x\in \mathscr{H}$. Prove that there exists $y\in\mathscr{H}$ and subsequence $\{S_{k_n}\}$ of $\{S_n\}$ such that for any $u\in\mathscr{H}$ we have $$\lim_{n\rightarrow\infty} \langle S_{k_n}u,x\rangle=\langle u,y\rangle$$ using Banach-Alaoglu Throrem

I've read the theorem multiple times, but I'm having trouble proving this.

@Mike Yes, it is Hilbert space. I will edit to add the details. — user640718, Apr 21 '21 at 14:12
(1) I think you meant $S_{k_n}$ in the centered equation. (2) What is ${S_n}$? — cmk, Apr 21 '21 at 14:14
@cmk I edited the question including all the details. I got carried away with the problem itself and I forgot to do so. — user640718, Apr 21 '21 at 14:20

folouer of kaklas · Accepted Answer · 2021-04-21T14:49:27.093

Since the $S_{n}$ are bounded, the adjoints $S_{n}^{*}$ exist and are bounded. In particular, $\sup_{n}\|S_{n}^{*}\|=\sup_{n}\|S_{n}\|= M$. For fixed $x \in \mathscr{H}$, define the sequence $x_{n}=S_{n}^{*}x$. This sequence is bounded; indeed:

$\|x_{n}\|\leq \|S_{n}^{*}\|\|x\|\leq M \|x\|$

Since $\mathscr{H}$ is a Hilbert space, the Banach Alaoglu theorem implies that every bounded sequence in it has a weakly convergent subsequence. Written out, this means that there exists a subsequence $x_{k_{n}}$ and $y\in \mathscr{H}$ such that for every $u \in \mathscr{H}$,

$\lim_{n\rightarrow\infty} \langle u,x_{k_{n}}\rangle=\langle u,y\rangle$.

By the properties of the adjoint, this can be rewritten as

$\lim_{n\rightarrow\infty} \langle S_{k_{n}}u,x\rangle=\langle u,y\rangle$.

Prove that $\lim_{n\rightarrow\infty} \langle S_{k_n}u,x\rangle=\langle u,y\rangle$ using Banach-Alaoglu Throrem

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